Answer:
0.75M Fe²⁺
Explanation:
First, we need to balance the redox reaction in acidic medium. Then, we can obtain moles of KMnO4 and with the reaction moles and molarity of the Fe²⁺ solution:
<em>Redox Balance:</em>
Fe²⁺ → Fe³⁺ + 1e⁻
5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
___________________________
5Fe²⁺ + 5e⁻ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
<h3>5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O</h3>
<em>Moles of KMnO₄:</em>
70.0mL = 0.0700L * (0.150mol / L) = 0.0105 moles KMnO₄
<em>Moles and molarity Fe²⁺:</em>
0.0105 moles KMnO₄ * (5 moles Fe²⁺ / 1mol KMnO₄) = 0.0525 moles Fe²⁺
In 70.0mL = 0.0700L:
0.0525 moles Fe²⁺ / 0.0700L =
<h3>0.75M Fe²⁺</h3>
I think it’s the second one!! Hope this helps! BRANLIEST plzzzz
Answer:
A. 0.641 L
Explanation:
<em>Given data:</em>
V1 = 2.5 L
T1 = 300.0 K
T2= 77.0 K
<em>To find: </em>
V2= ?
<em>Formula :</em>
By using Charles’ Law
<em>Calculation:</em>
V2 = 2.5 L x 77.0 K / 300.0 K
= 192.5 / 300.0 k
V2 = 0.641 L
The molecular weight of K2SO4 is 174.26 g/mole. The mass of K2SO4 required to make this solution is calculated in the following way.
550mL * (0.76mole/1000mL) * (174.26g/mole) = 72.84gram
<span>I hope this helps.</span>